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Question: Let $A$, $B$, $C$ be groups. Assume that $A \cong B$ and $B\cong C$. Prove that $A \cong C$.

Proof:

Let $\phi: A \to B$ and $\sigma: B \to C$

Suppose $\sigma \circ \phi: A \to C$

$i). One-to-one:$

Then $\sigma \circ \phi$ is one-to-one. Let $a, b \in A$

Suppose $(\sigma \circ \phi)(a) = (\sigma \circ \phi)(b)$

$\Rightarrow$ $\sigma (\phi(a)) = \sigma (\phi(b))$

(Since $\sigma$ is one-to-one.)

$\Rightarrow$ $(\phi(a)) = (\phi(b))$

(Since $\phi$ is one-to-one.)

$\Rightarrow$ $a = b$

$ii). Onto:$

Then $\sigma \circ \phi$ is onto.

Let $c \in C$. Because $\sigma$ is onto there exists $b \in B$ such that $\sigma(b) = c$. Also since $\phi$ is onto there exists a $a \in A$ such that $\phi(a) = b$.

Then $\sigma(\phi(a)) = \sigma(b) = c$

Then $\sigma \circ \phi$ is onto.

$iii).Opreation - Perserving:$

Let $x_1, x_2 \in A$.

Then $(\sigma \circ \phi) = \sigma(\phi(x_1, x_2)) = \sigma(\phi(x_1)\phi(x_2))$

Because $\phi(x_1) \in B$ and $\phi(x_2) \in B$,

$\sigma(\phi(x_1)\sigma(\phi(x_2)) = (\sigma \circ \phi)(x_1)(\sigma \circ \phi)(x_2)$

Then $\sigma \circ \phi$ is operation-preserving.

$\therefore A \cong C$

I know it isn't the most elegant proof, but I was wondering if this is totally correct or if I missed something or did something incorrectly. Any help appreciated!

Trever
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  • There are a number of typos in your proof of operation preserving I believe, but the proof overall looks very reasonable to me. Notice that en route you've proven that the composition of isomorphisms is also an isomorphism. – theyaoster Oct 19 '18 at 04:14
  • Yeah that part is confusing to me. Can you tell me what I should change? Thanks for the response @theyaoster . – Trever Oct 19 '18 at 04:22
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    Your second line should probably start with "Then $(\sigma \circ \phi)(x_1x_2) = \sigma(\phi(x_1x_2))$..." and your fourth line should probably start with "$\sigma(\phi(x_1)\phi(x_2)) = \sigma(\phi(x_1))\sigma(\phi(x_2))$..." and I think that would clear things up. – theyaoster Oct 19 '18 at 04:29

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